High-frequency AC transformer output hooked to half-wave DC voltage doubler. Current through is 6 A to LOW frequency switching load.
I assumed there was enough 20 kHz time to keep the second cap charged, ready for discharge at low frequency. The first cap is hot, the second one is cold, using camera photo. Fast-discharge caps 47 μF for lower ESR, but not low-ESR capacitors.
Is there something wrong with the first cap, or is there misalignment with the reactance 12 Ω of the secondary of the transformer?
Another method is resonance of Xl = Xc to have maximum energy transfer with cap in series. I don't know. The formula for matching says 0.6 μF, but the formula for a filter capacitor for a 6 ADC capable flow says 47 μF.
I don't want to burn out my power supply. With short time on and off of power supply, I get 6 A, but the first capacitor of the voltage doubler is hot. Should I put 0.6 μF as the first cap and 47 μf as the second cap of the voltage doubbler?
The 20 kHz is the frequency of inverter secondary that is charging the voltage doubler. This circuit is unregulated. The 20 kHz AC is applied to the doubler circuit as frequency pulses. The second cap gets 38 VDC. Camera circuit discharge caps were used instead of common low-ESR caps, which have less ESR than standard caps, but not as low a resistance as low-ESR caps.
A voltage doubler is called first stage, and has a first and a second following capacitor. Is this the formula? C farad = I current divided by frequency hertz times (2 * Epeak - E average volts). Peak volts = E average volts times square root of 2.
